Question: Evaluate $\int\dfrac{\cos4x}{2+\sin4x}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14\ln|2+\cos4x|+C$ (Choice B) B $\ln|2+\sin4x|+C$ (Choice C) C $\dfrac14\ln|2+\sin4x|+C$ (Choice D) D $\dfrac14\ln|\cos4x|+C$
Explanation: Note that the numerator of the integrand is almost the derivative of the denominator. We use a $~u$ -substitution to make things clearer. Let $~u=2+\sin4x\,$. Then $~du=4\cos4x\,dx\,$. That gives us $~~\dfrac14du=\cos4x\,dx\,$. We now change variables using this substitution. $ \int\dfrac{\cos4x}{2+\sin4x}\,dx~~~~~\Rightarrow~~~~~~~\dfrac14\int\dfrac1u\,du =\dfrac14\ln|u|+C$ Now substitute back $~(\,2+\sin 4x~$ in place of $~u)~$ to obtain the result in terms of $~x\,$. $ \dfrac14\ln|u|+C~~~~~\Rightarrow~~~~~\int\dfrac{\cos4x}{2+\sin4x}\,dx=\dfrac14\ln|2+\sin4x|+C$